To calculate the volume of the region limited by the cylinder with $$\ x^2+y^2=1 \ $$ and the planes $$\ z=0 \ $$ and $$ \ z=2-x$$.
Development:
For the symmetry of the problem, we will consider cylindrical coordinates.
The cylinder remains determined by $$r^2=1$$ and the planes $$z=0$$ and $$z=2-\cos\varphi$$. Therefore, the integration limits will be:
$$\begin{array}{l} r\in(0,1) \\ \varphi\in(0,2\pi) \\ z\in(0,2-r\cdot\cos\varphi) \end{array} $$
Then, if $$V$$ is the region under consideration:
$$$\begin{array}{rl} \text{Vol}(V)=&\iiint_V 1 \ dx \ dy \ dz =\int_0^{2\pi} \int_0^1 \int_0^{2-r\cos\varphi} r \ dz \ dr \ d\theta \\ =&\int_0^{2\pi} \int_0^1 \big[ r\cdot z \big]_0^{2-r\cos\varphi} \ dr \ d\theta = \int_0^{2\pi} \int_0^1 (2-r\cos\varphi)r \ dr \ d\theta \\ =& \int_0^{2\pi} \int_0^1 (2r-r^2\cos\varphi) \ dr \ d\varphi = \int_0^{2\pi} \Big[r^2-\dfrac{r^3}{3}\cos\varphi\Big]_0^1 \ dr \ d\varphi \\ =& \int_0^{2\pi} 1-\dfrac{1}{3}\cos\varphi \ d\varphi = \Big[\varphi-\dfrac{1}{3}\sin\varphi\Big]_0^{2\pi} =2\pi \end{array}$$$
Solution:
$$\text{Vol}(V)=2\pi$$